UVa 11988 Broken Keyboard (a.k.a. Beiju Text)(链表)

You're typing a long text with a broken keyboard. Well it's not so badly broken. The only problem with the keyboard is that sometimes the "home" key or the "end" key gets automatically pressed (internally).

You're not aware of this issue, since you're focusing on the text and did not even turn on the monitor! After you finished typing, you can see a text on the screen (if you turn on the monitor).
In Chinese, we can call it Beiju. Your task is to find the Beiju text.

Input

There are several test cases. Each test case is a single line containing at least one and at most 100,000 letters, underscores and two special characters '[' and ']'. '[' means the "Home" key is pressed internally, and ']' means the "End" key is pressed internally. The input is terminated by end-of-file (EOF). The size of input file does not exceed 5MB.

Output

For each case, print the Beiju text on the screen.

Sample Input

This_is_a_[Beiju]_text
[[]][][]Happy_Birthday_to_Tsinghua_University

Sample Output

BeijuThis_is_a__text
Happy_Birthday_to_Tsinghua_University

题意

每次把“[”和"]"里面的东西放到最前面

题解

正常暴力不用考虑，TLE

数组操作的就不说了，这是用链表做的

代码

#include<bits/stdc++.h>
using namespace std;
struct node
{
    char data;
    struct node *next;
};
int main()
{
    char s[100005];
    while(scanf("%s",s)!=EOF)
    {
        node *head=(node*)malloc(sizeof(node)),*rear,*p;
        head->next=NULL;
        p=rear=head;//第一个对头操作
        for(int i=0;s[i];i++)
        {
            char ch=s[i];
            if(ch=='[')
                p=head;
            else if(ch==']')
                p=rear;
            else
            {
                node *q=(node*)malloc(sizeof(node));
                q->data=ch;
                q->next=p->next;
                p->next=q;
                if(p==rear)
                    rear=q;
                p=q;
            }
        }
        p=head->next;
        while(p)
        {
            printf("%c",p->data);
            p=p->next;
        }
        printf("
");
    }
    return 0;
}